NIMCET 2023 — Mathematics PYQ

Q: How do I solve NIMCET 2023 Mathematics questions?

Practice NIMCET 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

If $\prod_{i = 1}^{n} tan (α_{i}) = 1 \forall α_{i} \in [0, \frac{π}{2}]$ Where $i = 1, 2, 3, \dots, n$ . Then, maximum value of $\prod_{i = 1}^{n} sin α_{i}$

Choose the correct answer:

Explanation

$ sin α_{1} sin α_{2} sin α_{3} \dots sin α_{n} = cos α_{1} cos α_{2} \dots cos α_{n} $

Multiplying both sides by
$ sin α_{1} sin α_{2} sin α_{3} \dots sin α_{n}, $
we get
$ sin^{2} α_{1} sin^{2} α_{2} sin^{2} α_{3} \dots sin^{2} α_{n} = (sin α_{1} cos α_{1}) (sin α_{2} cos α_{2}) \dots (sin α_{n} cos α_{n}) $

Using the identity
$ sin θ cos θ = \frac{1}{2} sin (2 θ), $

we obtain
$ sin^{2} α_{1} sin^{2} α_{2} sin^{2} α_{3} \dots sin^{2} α_{n} = \frac{1}{2 ^{n}} sin (2 α_{1}) sin (2 α_{2}) \dots sin (2 α_{n}) $

Since the maximum value of $sin θ$ is $1$ ,
$ sin^{2} α_{1} sin^{2} α_{2} \dots sin^{2} α_{n} \leq \frac{1}{2 ^{n}} $

Taking square root:
$ i = 1 \prod n sin α_{i} \leq \frac{1}{2 ^{n /2}} $

Hence,
$ Maximum value of i = 1 \prod n sin α_{i} = \frac{1}{2 ^{n /2}} $

Choose the correct answer:

Explore More